7.3 Pressure and Volume (Boyle’s Law) Chapter 7 Gases 7.3 Pressure and Volume (Boyle’s Law)

Boyle’s Law Boyle’s law states that the pressure of a gas is inversely related to its volume when T and n are constant if the pressure (P) increases, then the volume (V) decreases 2

PV Constant in Boyle’s Law The product P x V is constant as long as T and n do not change. P1V1 = 8.0 atm x 2.0 L = 16 atm L P2V2 = 4.0 atm x 4.0 L = 16 atm L P3V3 = 2.0 atm x 8.0 L = 16 atm L Boyle’s law can be stated as P1V1 = P2V2 (T, n constant)

Solving for a Gas Law Factor The equation for Boyle’s law can be rearranged to solve for any factor. P1V1 = P2V2 Boyle’s Law To solve for V2 , divide both sides by P2. P1V1 = P2V2 P2 P2 V1 x P1 = V2 P2

Boyle’s Law and Breathing: Inhalation During inhalation, the lungs expand the pressure in the lungs decreases air flows towards the lower pressure in the lungs

Boyle’s Law and Breathing: Exhalation During exhalation, lung volume decreases pressure within the lungs increases air flows from the higher pressure in the lungs to the outside

Guide to Calculations with Gas Laws

Calculation with Boyle’s Law Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of an 8.0 L sample of Freon gas after its pressure is changed from 550 mmHg to 2200 mmHg at constant T? STEP 1 Set up a data table: Conditions 1 Conditions 2 Know Predict P1 = 550 mmHg P2 = 2200 mmHg P increases V1 = 8.0 L V2 = ? V decreases

Calculation with Boyle’s Law (continued) STEP 2 Solve Boyle’s law for V2. When pressure increases, volume decreases. P1V1 = P2V2 V2 = V1 x P1 P2 STEP 3 Set up problem V2 = 8.0 L x 550 mmHg = 2.0 L 2200 mmHg pressure ratio decreases volume

Learning Check For a cylinder containing helium gas, indicate if cylinder A or cylinder B represents the new volume for the following changes (n and T are constant): 1) Pressure decreases 2) Pressure increases

Solution For a cylinder containing helium gas, indicate if cylinder A or cylinder B represents the new volume for the following changes (n and T are constant): 1) Pressure decreases (cylinder B) 2) Pressure increases (cylinder A)

Learning Check If a sample of helium gas has a volume of 120 mL and a pressure of 850 mmHg, what is the new volume if the pressure is changed to 425 mmHg? 1) 60 mL 2) 120 mL 3) 240 mL

Solution Conditions 1 Conditions 2 Know Predict 3) 240 mL Conditions 1 Conditions 2 Know Predict P1 = 850 mmHg P2 = 425 mmHg P decreases V1 = 120 mL V2 = ? V increases V2 = V1 x P1 = 120 mL x 850 mmHg = 240 mL P2 425 mmHg Pressure ratio increases volume

Learning Check A sample of helium gas in a balloon has a volume of 10. L at a pressure of 0.90 atm. At 1.40 atm (T constant), is the new volume represented by A, B, or C?

Solution A sample of helium gas in a balloon has a volume of 10. L at a pressure of 0.90 atm. At a higher pressure (T constant), the new volume is represented by the smaller balloon.

Learning Check A) 3.2 L B) 6.4 L C) 12.8 L If the sample of helium gas has a volume of 6.4 L at a pressure of 0.70 atm, what is the new volume when the pressure is increased to 1.40 atm (T constant)? A) 3.2 L B) 6.4 L C) 12.8 L

Solution STEP 1 Set up data table (conditions) A) 3.2 L STEP 1 Set up data table (conditions) Conditions 1 Conditions 2 Know Predict P1 = 0.70 atm P2 = 1.40 atm P increases V1 = 6.4 L V2 = ? V decreases STEP 2 Solve for Boyles law for V2 V2 = V1 x P1 P2 STEP 3 Set up problem V2 = 6.4 L x 0.70 atm = 3.2 L 1.40 atm Volume decreases when there is an increase in the pressure (temperature is constant).

Learning Check (T and n constant) 1) 200 mmHg 2) 400 mmHg 3) 1200 mmHg A sample of oxygen gas has a volume of 12.0 L at 600 mmHg. What is the new pressure when the volume changes to 36.0 L? (T and n constant) 1) 200 mmHg 2) 400 mmHg 3) 1200 mmHg

Solution P2 = P1 x V1 V2 600 mmHg x 12.0 L = 200 mmHg 36.0 L Conditions 1 Conditions 2 Know Predict P1 = 600 mmHg P2 = ? P decreases V1 = 12.0 L V2 = 36.0 L V increases P2 = P1 x V1 V2 600 mmHg x 12.0 L = 200 mmHg 36.0 L